Finding Normal Mode Frequencies for Two Hanging Masses with Springs

In summary, two masses (M_1 and M_2) suspended by springs (with constants k_1 and k_2 respectively) have equilibrium heights (y_1 and y_2). The Lagrangian (T-V) is being determined and the kinetic energy (T) for both masses is found. The potential energy for the second mass (V_2) is calculated to include a gravitational term and a spring term, while the potential energy for the first mass (V_1) only includes a gravitational term and a spring term, possibly with a positive or negative sign. The equations of motion are then used to find the normal mode frequencies, but the presence of gravitational terms makes it difficult to cancel off the exponentials
  • #1
Pacopag
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Homework Statement


A mass [tex]M_1[/tex] is suspended by a spring with constant [tex]k_1[/tex]. A second mass [tex]M_2[/tex] is suspended from the first by a spring with constant [tex]k_2[/tex]. The equilibrium height of the first is [tex]y_1[/tex] and the equilibrium height of the second is [tex]y_2[/tex].

I just want to write down the Lagrangian for now.


Homework Equations


[tex]L=T-V[/tex] (i.e. Lagrangian is Kinetic energy minus Potential energy).


The Attempt at a Solution


I'm using as generalized coordinates the displacements of the two masses from equilibrium [tex]\delta_1[/tex] and [tex]\delta_2[/tex]
I'm pretty sure that the kinetic energy parts are easy
[tex]T_1 = {1\over 2}M_1 \dot\delta_1^2[/tex] and [tex]T_2 = {1\over 2}M_2 \dot\delta_2^2[/tex]
Now for the potential energy for the second mass I think is
[tex]V_2 = -M_2 g (y_2+\delta_1+\delta_2) + {1\over 2}k_2(\delta_2-\delta_1)^2[/tex].
The first term is the gravitational part (taking the zero to be at the ceiling where the first spring is anchored) and the second term is the spring part.
But for the first mass, I'm tempted to write
[tex]V_1 = -M_1 g (y_1+\delta_1) + {1\over 2}k_1 \delta_1^2[/tex].
I think the spring part is wrong because [tex]M_1[/tex] is attached to spring 2, so there should be some contribution involving [tex]k_2[/tex]. This leads me to want to add another term like [tex]{1\over 2}k_2(\delta_2-\delta_1)^2[/tex], but I don't know if it should be positive or negative, and also I feel like I am 'double counting' the spring term in [tex]V_2[/tex]. For example, if spring 1 is streched and spring 2 is compressed, I expect [tex]V_1[/tex] to be increased by the compression of spring 2. I hope this is clear.
 
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  • #2
I've almost convinced myself that I should have
[tex]V1=-M_1 g(y_1+\delta_1)+{1\over 2}k_1\delta_1^2\pm{1\over 2}k_2(\delta_2-\delta_1)^2[/tex].
The sign is fuzzy to me, but thinking about the force (saying that down is negative), I think I should keep the minus. If this is correct, then I can get the equations of motion, which brings me to my real question: How do I find the normal mode frequencies? From similar examples, I've tried putting [tex]\delta_j = A_j e^{i\omega t}[/tex] (i.e. same frequency for both displacements (gen. coords)), into the equations of motion, then finding [tex]\omega[/tex] such that the solution exists. But, the gravitational terms don't allow us to cancel off the exponentials, which usually happens. I hope you can follow.
 
  • #3
Sorry. It should just be
[tex]V_1=-M_1 g(y_1+\delta_1)+{1\over 2}k_1\delta_1^2[/tex].
But I still don't know how to find the normal mode frequencies.
 

FAQ: Finding Normal Mode Frequencies for Two Hanging Masses with Springs

How does the mass of the objects affect the oscillation of the springs?

The mass of the objects affects the oscillation of the springs by changing the period of the oscillation. The larger the mass, the longer the period of oscillation will be.

What is the relationship between the spring constant and the frequency of the oscillation?

The spring constant and the frequency of the oscillation have a direct relationship. As the spring constant increases, the frequency of the oscillation also increases.

How does the distance between the two masses affect the amplitude of the oscillation?

The distance between the two masses has a direct relationship with the amplitude of the oscillation. As the distance between the masses increases, the amplitude of the oscillation decreases.

How does the damping factor affect the oscillation of the springs?

The damping factor affects the oscillation of the springs by reducing the amplitude of the oscillation over time. A higher damping factor results in a quicker decrease in the amplitude of the oscillation.

What happens to the oscillation if one of the masses is significantly larger than the other?

If one of the masses is significantly larger than the other, the oscillation will be dominated by the larger mass. The smaller mass will have very little effect on the oscillation, resulting in a longer period and smaller amplitude.

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